Problem: Divide the following complex numbers. $\dfrac{-6+8i}{-4-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-4+3i}$. $ \dfrac{-6+8i}{-4-3i} = \dfrac{-6+8i}{-4-3i} \cdot \dfrac{{-4+3i}}{{-4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-6+8i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-6+8i) \cdot (-4+3i)} {(-4)^2 - (-3i)^2} $ $ = \dfrac{(-6+8i) \cdot (-4+3i)} {16 + 9} $ $ = \dfrac{(-6+8i) \cdot (-4+3i)} {25} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-6+8i}) \cdot ({-4+3i})} {25} $ $ = \dfrac{{-6} \cdot {(-4)} + {8} \cdot {(-4) i} + {-6} \cdot {3 i} + {8} \cdot {3 i^2}} {25} $ $ = \dfrac{24 - 32i - 18i + 24 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{24 - 32i - 18i - 24} {25} = \dfrac{0 - 50i} {25} = -2i $